将一个数表示为 k 个斐波那契数之和的方法数
给定两个数字 N 和 K，找出将 N 表示为 K 个斐波那契数之和的方法。 例 :
Input : n = 12, k = 1
Output : 0
Input : n = 13, k = 3
Output : 2
Explanation : 2 + 3 + 8, 3 + 5 + 5\.
逼近:斐波那契数列为 f(0)=1，f(1)=2，f(i)=f(i-1)+f(i-2)为 i > 1。让我们假设 F(x，k，n)是利用来自 f(0)，f(1)，…f(n-1)的恰好 k 个数形成和 x 的方法数。要找到 F(x，k，n)的一个递归，注意有两种情况:f(n-1)是否在和中。
- 如果 f(n-1)不在和中，则 x 是使用来自 f(0)、f(1)、…、f(n-2)的恰好 k 个数形成的和。
- 如果 f(n-1)在和中，则剩余的 x-f(n-1)是使用来自 f(0)、f(1)、…、f(n-1)的恰好 k-1 个数字形成的。(请注意，f(n-1)仍然包含在内，因为允许重复的数字。).
所以递推关系为:
F(x，k，n)= F(x，k，n-1)+F(x-f(n-1)，k-1，n)
基本情况:
- 如果 k=0，那么数列中有零个数字，所以和只能是 0。因此，F(0，0，n)=1。
- F(x，0，n)=0，如果 x 不等于 0。
此外，还有其他情况使 F(x，k，n)=0，如下所示:
- 如果 k>0 且 x=0，因为至少有一个正数必然导致正和。
- 如果 k>0 并且 n=0，因为没有可能选择剩下的数字。
- 如果 x<0，因为没有办法用有限数量的非负数形成负和。
以下是上述方法的实现:
C++
// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
// to store fibonacci numbers
// 42 second number in fibonacci series
// largest possible integer
int fib[43] = { 0 };
// Function to generate fibonacci series
void fibonacci()
{
fib[0] = 1;
fib[1] = 2;
for (int i = 2; i < 43; i++)
fib[i] = fib[i - 1] + fib[i - 2];
}
// Recursive function to return the
// number of ways
int rec(int x, int y, int last)
{
// base condition
if (y == 0) {
if (x == 0)
return 1;
return 0;
}
int sum = 0;
// for recursive function call
for (int i = last; i >= 0 and fib[i] * y >= x; i--) {
if (fib[i] > x)
continue;
sum += rec(x - fib[i], y - 1, i);
}
return sum;
}
// Driver code
int main()
{
fibonacci();
int n = 13, k = 3;
cout << "Possible ways are: "
<< rec(n, k, 42);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
//Java implementation of above approach
public class AQW {
//to store fibonacci numbers
//42 second number in fibonacci series
//largest possible integer
static int fib[] = new int[43];
//Function to generate fibonacci series
static void fibonacci()
{
fib[0] = 1;
fib[1] = 2;
for (int i = 2; i < 43; i++)
fib[i] = fib[i - 1] + fib[i - 2];
}
//Recursive function to return the
//number of ways
static int rec(int x, int y, int last)
{
// base condition
if (y == 0) {
if (x == 0)
return 1;
return 0;
}
int sum = 0;
// for recursive function call
for (int i = last; i >= 0 && fib[i] * y >= x; i--) {
if (fib[i] > x)
continue;
sum += rec(x - fib[i], y - 1, i);
}
return sum;
}
//Driver code
public static void main(String[] args) {
fibonacci();
int n = 13, k = 3;
System.out.println("Possible ways are: "+ rec(n, k, 42));
}
}
Python 3
# Python3 implementation of the above approach
# To store fibonacci numbers 42 second
# number in fibonacci series largest
# possible integer
fib = [0] * 43
# Function to generate fibonacci
# series
def fibonacci():
fib[0] = 1
fib[1] = 2
for i in range(2, 43):
fib[i] = fib[i - 1] + fib[i - 2]
# Recursive function to return the
# number of ways
def rec(x, y, last):
# base condition
if y == 0:
if x == 0:
return 1
return 0
Sum, i = 0, last
# for recursive function call
while i >= 0 and fib[i] * y >= x:
if fib[i] > x:
i -= 1
continue
Sum += rec(x - fib[i], y - 1, i)
i -= 1
return Sum
# Driver code
if __name__ == "__main__":
fibonacci()
n, k = 13, 3
print("Possible ways are:", rec(n, k, 42))
# This code is contributed
# by Rituraj Jain
C
// C# implementation of above approach
using System;
class GFG
{
// to store fibonacci numbers
// 42 second number in fibonacci series
// largest possible integer
static int[] fib = new int[43];
// Function to generate fibonacci series
public static void fibonacci()
{
fib[0] = 1;
fib[1] = 2;
for (int i = 2; i < 43; i++)
fib[i] = fib[i - 1] + fib[i - 2];
}
// Recursive function to return the
// number of ways
public static int rec(int x, int y, int last)
{
// base condition
if (y == 0) {
if (x == 0)
return 1;
return 0;
}
int sum = 0;
// for recursive function call
for (int i = last; i >= 0 && fib[i] * y >= x; i--) {
if (fib[i] > x)
continue;
sum += rec(x - fib[i], y - 1, i);
}
return sum;
}
// Driver code
static void Main()
{
for(int i = 0; i < 43; i++)
fib[i] = 0;
fibonacci();
int n = 13, k = 3;
Console.Write("Possible ways are: " + rec(n, k, 42));
}
//This code is contributed by DrRoot_
}
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of above approach
// To store fibonacci numbers
// 42 second number in fibonacci series
// largest possible integer
$fib = array_fill(0, 43, 0);
// Function to generate
// fibonacci series
function fibonacci()
{
global $fib;
$fib[0] = 1;
$fib[1] = 2;
for ($i = 2; $i < 43; $i++)
$fib[$i] = $fib[$i - 1] +
$fib[$i - 2];
}
// Recursive function to return
// the number of ways
function rec($x, $y, $last)
{
global $fib;
// base condition
if ($y == 0)
{
if ($x == 0)
return 1;
return 0;
}
$sum = 0;
// for recursive function call
for ($i = $last; $i >= 0 and
$fib[$i] * $y >= $x; $i--)
{
if ($fib[$i] > $x)
continue;
$sum += rec($x - $fib[$i],
$y - 1, $i);
}
return $sum;
}
// Driver code
fibonacci();
$n = 13;
$k = 3;
echo "Possible ways are: " .
rec($n, $k, 42);
// This code is contributed by mits
?>
java 描述语言
<script>
//Javascript implementation of above approach
//to store fibonacci numbers
//42 second number in fibonacci series
//largest possible integer
let fib=new Array(43);
//Function to generate fibonacci series
function fibonacci()
{
fib[0] = 1;
fib[1] = 2;
for (let i = 2; i < 43; i++)
fib[i] = fib[i - 1] + fib[i - 2];
}
//Recursive function to return the
//number of ways
function rec(x,y,last)
{
// base condition
if (y == 0) {
if (x == 0)
return 1;
return 0;
}
let sum = 0;
// for recursive function call
for (let i = last; i >= 0 && fib[i] * y >= x; i--) {
if (fib[i] > x)
continue;
sum += rec(x - fib[i], y - 1, i);
}
return sum;
}
//Driver code
fibonacci();
let n = 13, k = 3;
document.write("Possible ways are: "+ rec(n, k, 42));
// This code is contributed by rag2127
</script>
Output:
Possible ways are: 2
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